package main

import "fmt"

/*
Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it.
That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].
Return the answer in an array.

Example 1:
Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation:
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1).
For nums[3]=2 there exist one smaller number than it (1).
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:
Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Example 3:
Input: nums = [7,7,7,7]
Output: [0,0,0,0]

Constraints:
2 <= nums.length <= 500
0 <= nums[i] <= 100
*/

func main() {
	fmt.Println(smallerNumbersThanCurrent([]int{8, 1, 2, 2, 3}))
}

func smallerNumbersThanCurrent(nums []int) []int {
	ret := make([]int, len(nums))

	for idx, v := range nums {
		for _, v2 := range nums {
			if v > v2 {
				ret[idx]++
			}
		}
	}

	return ret
}
